Calculate the volume (in liters) of CO2 generated from 0.350 g of NaHCO3 and excess HCl(aq) at 1.00 atm and 37.0 °C.?

Molar mass of NaHCO₃ = (23.0 + 1.0 + 12.0 + 16.0×3) g/mol = 84.0 g/mol
No. of moles of NaHCO₃ reacted = (0.350 g) / (84.0 g/mol) = 0.00417 mol

NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g)
Mole ratio NaHCO₃ : CO₂ = 1 : 1
No. of moles of CO₂ generated = (0.00417 mol) × 1 = 0.00417 mol

For the CO₂ generated :
Pressure, P = 1.00 atm
Volume, V = ? L
No. of moles, n = 0.00417 mol
Gas constant, R = 0.00821 L atm / (mol K)
Temperature, T = (273 + 37) K = 310 K

For a fixed amount of gas: PV = nRT
Then, V = nRT/P

Volume of CO₂ formed = 0.00417 × 0.0821 × 310 / 1.00 L = 0.106 L